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Home , Bond order, Molecular orbital, Molecular orbital theory

Approach to Bonding

Chemistry 362; spring 2016 Marcetta Y. Darensbourg, Professor Xuemei Yang, Graduate Assistant Pokhraj Ghosh, Graduate Assistant

The following slides were mainly a gift from Professor Martyn Poliakoff Of the Department of in Nottingham, England. Tom Poliakoff also used these slides and prepared them, to my knowledge. You might also check The MIT open courseware lecture to refresh your memory of molecular orbitals. MIT Open Courseware lecture https://www.youtube.com/watch?v=llaa-iEYDLI HF https://www.youtube.com/watch?v=estiedAlXII https://www.youtube.com/watch?v=GD5CrjyAKx4 B2H6

http://www.chemtube3d.com/orbitalsCO.htm CO

Valence Bond Approach: Localized Bonds, just like Lewis Structures and Basis of VVSEPRB approa c h: overlap orbitals in each bond separately. Each bond is LOCALISED between two .

MOLECULAR ORBITAL APPROACH

In molecular orbital (MO) approach - overlap orbitals for the whole - bonding is therefore DELOCALISED. We will look first at DIATOMIC and only later move on to POLYATOMIC MOLECULES.

MOLECULAR ORBITAL FOR DIATOMIC MOLECULES

In principle, set up Schrödinger wave equation for molecule and solve it. Solution will involve molecular orbitals - similar to atomic orbitals - but centred around all of the nuclei in molecule. Each defined by sets of quantum numbers, with probability density determined by ψ2, where ψ = molecular .

Approximate method: At any moment, electron near one nucleus - approximate behaviour like electron in for that . Over time - electron associated with other nuclei in molecule. Therefore construct molecular orbitals (m.o.'s) by forming:

Linear Combination of Atomic Orbitals It is this, LCAO, method which we will use to construct m.o's.

Simplest example - H2: two H atoms HA and HB

Only two a.o.'s (1sA, 1sB) to form linear combinations.

General rule: n a.o.'s n m.o.'s

So we can only construct 2 m.o.'s for H2 - and these are:

ψb = 1sA + 1sB and ψa = 1sA - 1sB

i.e. the sum (ψb) and the difference (ψa) of the constituent a.o.'s.

Consider the electron distribution in each of these: Two non- interacting + + H atoms

Signs refer to sign of ψ ψ b = 1sA + 1sB +

+ NODE

+ –

ψ = 1s - 1s a A B Atom A Atom B Consider in each case the INTERNUCLEAR REGION

Probability of finding electron there is:

ψb > 1sA, 1sB > ψa

Electron in this region attracted to BOTH nuclei, therefore most favourable position. Hence, electron in ψb will be at lower energy than in non-interacting a.o.'s, and electron in ψa will be at higher energy still.

Thus an electron in ψb will hold the nuclei together, one in ψa will push them apart.

ψb is a BONDING m.o.,

ψa is an ANTI-BONDING m.o. Thus we can draw DIAGRAM for m.o.'s of H2 : ψa

1sA 1sB

ψb

HA H2 HB

By aufbau & Pauli principles - the 2 go into ψb - with paired spins. MO’s for H2 molecule

By Lewis/V.B. theory - one pair of electrons = one bond. To be consistent, in M.O. theory, define BOND ORDER as follows: Bond order = [(No. of electrons in bonding m.o.'s) – (No. of electrons in antibonding m.o.'s)]/2

Thus, for H2, bond order = (2 - 0)/2 = 1 (i.e. a - as expected)

Magnetic Properties of Molecules

All electrons paired - repelled by magnetic field - DIAMAGNETIC

One or more unpaired electrons - attracted into magnetic field - PARAMAGNETIC

H2 is diamagnetic. HETERONUCLEAR DIATOMIC MOLECULES

Simplest would be HHe. Differs from H2 in two ways: (1) A.O. energies for H, He different. He - greater nuclear charge, electrons more tightly bound. (2) Now three electrons to feed into m.o.'s.

Energy level diagram is now: ψa

1sH

average energy 1sHe of a.o.'s ψb

H HHe He

For heteronuclear diatomics, m.o.'s formed symmetrically above and below AVERAGE energy of constituent a.o.'s For HHe, bond order = (2 - 1)/2 = 1/2 i.e. v. wk. "1/2" bond - not formed under normal conditions - v. unstable.

Unpaired electron, PARAMAGNETIC.

Note for "He2" - extra electron in antibonding m.o. - therefore bond order = 0. Molecule does not exist - no force to hold atoms together. He is monatomic gas. M.O.'s for homonuclear diatomics (A2) for elements of first row of the Periodic Table

For Li2, Be2, B2 etc., more complex than for H2, HHe - more available a.o.'s - 1s, 2s, 2p. Are there restrictions on overlap?

(1) electrons only - core electrons too close to nucleus, too tightly bound

(2) Most efficient overlap between orbitals of same energy, i.e. for homonuclear diatomics this means 2s/2s, 2p/2p (for heteronuclear diatomics - see later)

(3) SYMMETRY RESTRICTIONS These are best shown pictorially

Let us see how this works for 2s and 2p orbitals. s/s + + . + . σ2s

s + s overlap everywhere positive→ BONDING M.O.

+ – + . . – σ*2s s – s overlap everywhere negative→ ANTI-BONDING M.O. For p orbitals - three per atom. Define z-axis as molecular axis. Hence pz orbitals can overlap in same way as s orbitals.

σ2pz pz/pz – + + – – . + . –

pz + pz overlap everywhere positive→ BONDING M.O.

– + – + – . + – . +

σ*2pz pz – pz overlap everywhere negative→ ANTI-BONDING M.O. px, py orbitals are perpendicular to axis, but can still interact

px/px + + + or . . π2px or π py/py – – – 2py

px + px overlap everywhere positive→ BONDING M.O. π + – + – *2px . . or π – + – + *2py px – px overlap everywhere negative→ ANTI-BONDING

M.O. Also exactly analogous pair from py. Need to consider all possibilities (could be needed for heteronuclear diatomics) px/px + + + or . . π2px or π py/py – – – 2py

π + – + – *2px . . or π – + – + *2py s/pz gives bonding and anti-bonding pair.

s/p σ z + + – + . . – spz

+ – + + . – . + σ*spz s/px or py and pz/px or py all non-bonding (positive and negative overlaps cancel. No overlap at all for px/py.

+ pz/px + s/px or non-bonding + non-bonding – + or pz/py – s/py –

Before moving on to show the energy level diagram for A2 molecules - we need to be clear about the labels for m.o.’s LABELS FOR M.O.'S

Two types of m.o. - in terms of symmetry to rotation about molecular axis.

s/s, pz/pz. s/pz - completely symmetrical to such rotation. All such given Greek symbol : σ ("sigma") o px/px, py/py - change sign every 180 rotation - these are

given symbol : π ("pi")

Note - same symbols as for valence bonds (above)

Bonding and antibonding orbitals of each type - differentiated by asterisk ("star") on the antibonding ones.

What is the m.o. energy level for these A2 molecules? M.O.Energy Level σ*2s Diagram for A2 (A = Li, Be) 2s 2s

σ2s Remember: 1s orbitals effectively non-bonding, A A2 A

Use Aufbau, Pauli, Hund - just as in filling atomic orbitals

2 0 Li2 Only two valence electrons, i.e. σs σ*s . Bond order = 1. Diamagnetic Li2 exists in gas phase over metallic lithium.

2 2 Bond order = 0 - no net bonding "Be2" σs σ*s energy, so molecule does not exist. Beryllium in gas phase is monatomic. m.o.'s derived from 2p a.o.'s:

σ*2pz

π*2px,π*2py

2p 2p

π2px,π2py

σ2pz

1. px, py alike in all respects except orientation, so m.o.'s derived from them must be degenerate.

2. sideways (π) overlap is less efficient than end-on (σ), so π m.o.'s less bonding than σ σ*2pz

M.O.Energy Level π*2px,π*2py Diagram for A2 (A = O ) 2p 2p

π2px,π2py

σ2pz

σ*2s

2s 2s

σ2s

A A2 A

O2 2 2 2 2 2 1 1 Electronic configuration: σs σ*s σpz πpx πpy π*px π*py Note Hund's rule again! Bond order = (8 - 4)/2 = 2 () and PARAMAGNETIC. V.B. theory could not explain .

O2 is paramagnetic O2 is only one of a series of diatomic species:

+ + – O2 - "oxidation of oxygen" in O2 PtF6 ; O2 - normal form of oxygen – + – 2– 2+ 2– O2 - superoxide ion, e.g. K O2 ; O2 - peroxide ion, e.g. Na2 O2

Electron configurations: + 2 2 2 2 2 1 0 O2 σs σ*s σpz πpx πpy π*px π*py elecron lost from anti-bonding m.o. - stronger bond than O2. Bond order = 2.5 Paramagnetic - but less so than O2 - only one unpaired electron. – σ 2σ 2σ 2π 2π 2π 2π 1 O2 s *s pz px py *px *py electron gained in anti-bonding m.o. - weaker bond than O2. Bond order = 1.5 Paramagnetic (one unpaired electron). Superoxide Dismutase (SOD)

2– 2 2 2 2 2 2 2 O2 σs σ*s σpz πpx πpy π*px π*py Bond order = 1. Diamagnetic Note bond lengths + O2 - 112 pm Weaker bonds, therefore O - 120 pm 2 longer bonds – O2 - 126 pm 2– O2 - 149 pm

Electronic configuration: 2– F2 Isoelectronic with O2 2 2 2 2 2 2 2 σs σ*s σpz πpx πpy π*px π*py Bond order = 1. Diamagnetic

2 2 2 2 2 2 2 2 "Ne2" σs σ*s σpz πpx πpy π*px π*py σ*pz Bond order = 0

Therefore does not exist. Neon is a monatomic gas

What about B, C and N??

Energy difference between s and p orbitals is not so great as in O and F. Therefore there is significant interaction of sigma molecular orbitals Derived from 2s and 2p.

Valence State Ionization Energies (eV) 2s 2p Δ

B 14.0 8.3 5.7 C 19.4 10.6 8.8 N 25.6 13.2 12.4 O 32.3 15.8 16.5 F 40.2 21.6 18.6

This leads to a different ordering of the mo‘s. The so-called “s-p mixing”. It’s not difficult – look it up in an advanced book, if you are interested What about B, C and N??

This is of major importance: N2 2 2 2 2 2 σs σ*s πpx πpy σpz

Bond order = (8 - 2)/2 = 3 i.e. , very strong, stable molecule. Diamagnetic. (Note the different order of s/p orbitals)

B and C 2 2 C2 is important in flames and comets, interstellar space

Light Emission from Comets Heteronuclear Diatomic Molecules (AB) Involving First Row Elements

Remember two basic principles:

(1) A.o. energies decrease as nuclear charge increases.

(2) Bonding/antibonding m.o.'s formed symmetrically below/above average of a.o. energies, respectively.

M.o. energy level diagram based on that for A2: Molecular Orbital Energy Level Diagram for a Heteronuclear Diatomic e.g. for CO - similar to N2 σ*2pz - but with different a.o. π*2px,π*2py energies for C and O, 2p i.e. O > C. 2p

Electronic configuration: π2px,π2py 2 2 2 2 2 σ2pz σs σ*s σpz πpx πpy σ*2s Bond order = (8 - 2)/2 = 3 2s

2s σ2s Just like N2 : ISOELECTRONIC C CO O (i.e. same no. of electrons). DIAMAGNETIC. A different type of heteronuclear diatomic: 1 2 2 2 2 1 HF H 1s F 1s 2s 2px 2py 2pz Overlap between H1s and F2pz giving:

+ . – . + + . . – and node σ σspz *spz (bonding) (anti-bonding) σ*spz

1s

2p

σspz 2s

H HF F

non-bonding electrons on F Electronic configuration: 2 4 2 0 [F2s 2p ]σsp σ*sp

Diamagnetic. Bond order = 1

F2s non-bonding - too low in energy

F2px,2py non-bonding because of wrong symmetry Molecular Orbitals in Polyatomic Molecules

In the vast majority of cases, bonds can be described as LOCALISED between pairs of nuclei - therefore can use VB approach.

At this stage we only need to use molecular orbitals (m.o’s) when DELOCALISATION of electrons occurs - i.e. when several Lewis structures have to be drawn.

Even in these cases we can use the VB approach for the σ-framework, and construct m.o.’s only for the π-bonds.

2- Just look at TWO examples - one inorganic (CO3 ), one organic (, C6H6). 2- 0 Carbonate ion, CO3 . Regular trigonal planar, equal bonds, all angles 120 .

2 σ-framework: use sp hybrids on both C and O's. 2 electrons in each bond + lone pairs in remaining sp2 hybrids on O's = 18 electrons

+ + Total number of valence electrons = 4 (C) + – 18 (3 O's) + 2 (negative charges) = 24 – O–

+ Therefore have 6 electrons to put into π-type orbitals –C – ++ + – + – + –

– – – π O – O Which orbitals are available for -bonding?

+ +

2 On each atom, sp hybrid formation uses px and py

orbitals. Therefore one pz orbital on each atom is available for π-bonding - 4 orbitals in total. z

+

+ +

O C + - - - O -

Remember general rule: n a.o's n m.o.'s

Therefore we will form 4 m.o.'s

Details of how to calculate what these look like left until later. By analogy with diatomic m.o.'s - most strongly bonding m.o. will be the one which increases electron density in all of the bonds:

+

+ +

O C + - - - O -

The most antibonding m.o. will have NODES on all of the bonds:

+ + -

O C +

- - + O -

nodes The remaining two turn out to be non-bonding, giving the 2- energy level for the π-bonding in CO3 as follows:

π∗ The 6 electrons just fill the bonding and non-bonding m.o.'s.

pz

0 π One π-bond shared out equally p z in a delocalised m.o.

π Each bond therefore 1σ + 1/3π

2- C CO3 3O

All electrons paired - therefore DIAMAGNETIC

Benzene, C6H6 Remember that there are two equivalent Lewis structures for benzene:

Simplified drawing - CH at each corner.

Benzene is in fact a regular hexagon - all C-C bonds equal, and with a length intermediate between C-C and C=C.

σ-framework - sp2 hybridisation at each C, forming a σ-bond to H and two adjacent C atoms

π-bonding - a pz orbital on each C, i.e. 6 a.o.’s, therefore 6 m.o.’s will be formed. + +

A.o.’s for π-bonding: + − − +

− + + −

− − sigma bonds in Benzene Pi-bonds C6H6 By analogy with carbonate - most strongly bonding m.o. formed by adding all of these a.o.’s together (incr. electron density in each bond).

Also by analogy with carbonate - most strongly antibonding m.o. formed by combining all of these a.o.’s together to give NODES between each pair of C atoms

In addition, the remaining 4 m.o.’s will comprise two WEAKLY BONDING and two WEAKLY ANTIBONDING combinations strongly antibonding Thus we have a π-bonding m.o. energy level for benzene: weakly antibonding

6 pz a.o.'s weakly bonding

strongly bonding Pi-bonds C6H6 There are 6 electrons (one per C atom) - and these fill the strongly and weakly antibonding m.o.’s. These orbitals are DELOCALISED, and explain the observed properties of benzene. Note stability as all bonding m.o.’s are full, and all electrons paired.

Note - there is a simple pictorial method for obtaining the relative energies of the π-m.o.’s of benzene and any other cyclic CnHn compound. Draw a circle round the shape of the molecule with one corner at the lowest part of the circle - and the corners just touching the circle. This gives the m.o. energies! Any molecule where π-electrons are delocalised is said to show CONJUGATION - and the properties of such molecules are different from those with localised multiple bonds -lots of examples in Organic Chemistry.

In particular, cyclic compounds like benzene (with a FULL SHELL of π-m.o’s) are described as being AROMATIC. confers specific properties on such compounds. In particular they are more stable (lower energy) than would have been predicted on the basis of localised double bond formation.